Republic of the Philippines

Department of Education
Region III
SCHOOLS DIVISION OF ZAMBALES
Zone 6, Iba, Zambales
Tel./Fax No. (047) 602 1391
E-mail Address: zambales@deped.gov.ph
website: www.depedzambales.ph

LEARNING ACTIVITY SHEET

PRE-CALCULUS
Science Technology Engineering and Mathematics (STEM) Specialized Subject
WEEK 1

I. Introduction
Welcome your first mathematics specialization subject in STEM,
Pre-Calculus. The first the topic in this subject is the conic sections (or
conics). Conic Sections are the curves that are formed when a
plane cuts a cone. When a cone is cut horizontally by a plane the
cross section is a circle. Circle is an example of a conic section.
Other result of cutting a cone can produce ellipses, parabolas and
hyperbolas. We can also consider line and a point as special cases
of conic sections.
As we explore conics, you will realise the applications in real life
that goes with the study sections. From parabolas that used in
satellite dishes, hyperbolas that is used designing telescope, ellipses
that model the orbits of the planets and circles, where you most
familiar with, that we use to design wheels in in engineering design
such as tunnels and arcs.
In this Learning Activity Sheet (LAS), we will focus on circle and
its degenerate form.
II. Learning Competencies

1
 Illustrate the different types of conic sections : parabola,
ellipse, circle, hyperbola, and degenerate cases.
(STEM_PC11AG-Ia-1)
Define a Circle (STEM_PC11AG-Ia-2) and determine the
Standard form of equation of a circle (STEM_PC11AG-Ia-2)
III. Objectives:
At the end of this learning activity sheet, you are
expected to:
1. Differentiate the different types of conic sections and
degenerate cases;
2. Appreciate and define circle
3. derive standard form and vertex form of circles from general
form;
4. graph circles and;
5. solve situational problems involving circles.

IV. Discussion

Conic Sections
A conic section (or simply conic) is a curve obtained as the
intersection of the surface of a cone with a plane. There are four
types of conic sections, the hyperbola, the parabola, the ellipse,
and the circle.

Conic sections can be generated by intersecting a plane with a

cone. A cone has two identically shaped parts called nappes. One
nappe is what most people mean by “cone,” and has the shape of
a party hat.

2
Figure 1. A cone and conic sections: The nappes and the four conic sections. Each
conic is determined by the angle the plane makes with the axis of the cone. (lifted from
courses.lumenlerning.com)

Looking closely at the figure above, the intersections between the

plane and the cone produces conics, here are the observations:
 When the plane cuts the cone horizontally, parallel to the x axis,
the resulting conic is a circle;
 when the plane cuts the one cone at an angle other from the x
axis, thus forming a bounded curve the result is an ellipse.
 when the plane cuts the one cone and form an unbounded
curve the result is a parabola,
 while when the planes cuts both cones and form an
unbounded curves on both curves the result is called
hyperbola.

3
 Below is the graph of four different conics on rectangular
coordinates.

Circle (x2+y2=9) Ellipse (2x2+y2=9)

Parabola (x2+y=3) Hyperbola (x2-y2=3)

Figure 2. Graph of Conic Sections

In some instances the intersection between the plane and the cones
doesn’t produce either of the above conic sections; this instance
produces what is called degenerate conics. A degenerate conic is
formed when a plane intersects the vertex of the cone. There are
three types of degenerate conics:
- The degenerate form of a circle or an ellipse is a singular point.
At the vertex of the cone, the radius is 0.
4
 - The degenerate form of a parabola is a line or two parallel
lines.
- The degenerate form of a hyperbola is two intersecting lines.

Fig 3. The Degenerate Conics; intersecting lines, a line and a point (lifted from
courses.lumenlerning.com)

The Definition and Equation of a Circle

A circle is formed when the plane cuts the cone exactly parallel to
its base. It can be said that the circle is a special kind of ellipse
(ellipse will be discussed in the next LAS).
The intersection with the cone is a set of points equidistant from a
common point (the central axis of the cone) and becomes the
center of the circle. Circles have certain features:
 A center point
 A radius, which the distance from any point on the circle to the
center point
5
On a coordinate plane, the general form of the equation of the
circle is (x−h)2+(y−k)2=r2 , where (h,k) are the coordinates of the
center of the circle, and r is the radius. A point is a degenerate form
of the circle that is formed when the single point intersection
happens. That is, when the plane only intersects the very tip of the
cone. The value of the radius is therefore zero. Further, the
eccentricity 𝒆 (is the ratio of the distance from the center to the foci
and the distance from the center to the vertices) of a circle is also
equal to zero. The essence of eccentricity will be fully discussed on
the LAS for ellipses.

Figure 4 (a) circle with a center at coordinate C at (3,1). (b) circle with a center at
coordinate C at (x,y)

Study figure 4a. Two points from a circle given are A (-2,1) and
B(6,5). By looking at graph the distance the distance of A from C is 5
units, while the distance of B from C can be calculated using
distance formula, 𝐵𝐶 = √(6 − 3)2 + (5 − 1)2 which is equal to 5, we
can then that distance PC is equal to 5. The collection of points that
is 5 units from C, forms a circle.

6
In figure 4b, it clearly shows that distance between point C and P is
called r, r stands for radius.

We can then say that, letting C be a given point. The set of all points
having the same distance from C is called a circle. Thus the point is
called a center of the circle and the common distance from any set
points in the circle from C is its radius.

Let us again consider figure 4b. The center of the circle is C (h, k)
while the radius is greater than zero r >0 (if the radius is equal to zero,
the figure becomes degenerate circle: a point). A point P(x,y) lies
on the circle if and only if PC = r ; hence we sat that .
𝑃𝐶 = 𝑟
by the distance formula √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 Equation 1

Equation 1 is what we call the Standard Form equation of a Circle

with the center (h, k) and having a radius of r. At the origin with
centre (0, 0) the standard equation is reduced to 𝑥 2 + 𝑦 2 = 𝑟 2 .
Let us study the following examples:
1. Center at the origin, and the radius is 3
2. Center (2,-3), radius of 4
3. The circle in Figure 4a
4. Circle A in Fig 5
5. Circle B in Fig 5

7
 6. Center (0,0) and the circle touches (0,3)
7. Center (0, -4) tangent to x axis
8. Center (5, -6) tangent to y axis
9. Center (-5, -5) tangent to both x and y axis
10. It has diameter with endpoints A(-1, 4) and B(4, 2)

Figure 5: example circles

Solutions.
(1) x2 + y2 = 9
(2) (x-2)2 + (y+3)2 = 16
(3) The centre is (3,1) and the radius is 5 so the equation is
(x-3)2 + (y-1)2 = 25
(4) The centre is (1,1) and the radius is 2 so the equation is
(x-1)2 + (y-1)2 = 4
(5) The centre is (-1,-1) and the radius is 3 so the equation is
(x+1)2 + (y+1)2 = 9

8
 (6) The centre is (0,0), the distance between (0,0) and (0,3) using
distance formula √(0 − 0)2 + (3 − 0)2 = 𝑟 , r= 3, so the
equation is x2 + y2 = 9
(7) The centre is (0,-4), tangent to x means it touches and (0,0)
using distance formula √(0 − 0)2 + (0 + 4)2 = 𝑟 , r= 4, so the
equation is x2 +(y+4)2 = 16
(8) The centre is (-5,6), tangent to y-axis means it touches (-5,0).
Using distance formula √(−5 + 5)2 + (0 − 6)2 = 𝑟 , r= 6, so the
equation is (x+5)2 +(y-6)2 = 36
(9) The centre is (-5,-5), tangent to x-axis and y-axis this means it
touches (-5,0) and (0-5). Using distance formula
√(−5 + 5)2 + (0 + 5)2 = 𝑟 , r= 5, so the equation is (x+5)2
+(y+5)2 = 25
−1+4 4+2
(10) The center C is the midpoint of A and B: C = ( , )=
2 2

3 3 2 29
(2 , 3) . The radius is then = 𝐴𝐶 = √(−1 − 3) + (4 − 3)2 = √ 4 .

3 2 29
The circle has the equation (𝑥 − ) + (𝑦 − 3)2 =
2 4

Standard and General form of the Equation of the Circle

Study example 2 from the previous exercise, we have obtain the
equation (x-2)2 + (y+3)2 = 16 this equation as we know it is in standard
form, if we expand of simplify the equation we write this as
𝑥 2 − 4𝑥 + 4 − 𝑦 2 + 6𝑦 + 9 = 16 further
𝑥 2 − 4𝑥 − 𝑦 2 + 6𝑥 − 3 = 0 re-arrangement give us
𝑥 2 − 𝑦 2 − 4𝑥 + 6𝑥 − 3 = 0 final equation
9
The final equation above is what we call equation of the circle in
General Form. If the equation of the circle is in general form then it is
written in this manner
𝐴𝑥 2 + 𝐴𝑦 2 + 𝐶𝑥 + 𝐷𝑦 + 𝐸 = 0 , hence 𝐴 ≠ 0
Note that circle has coefficient of 𝑥 2 and 𝑦 2 are equal, hence the
general form can be simply written as
𝑥 2 + 𝑦 2 + 𝐶𝑥 + 𝐷𝑦 + 𝐸 = 0
Equations are normally written in general form. However equation of
a circle in general form doesn’t say much about its properties; its
center and its radius are not defined. Hence converting general
form to standard form is a key skill to unlock the properties of a circle.
In the conversion process from general form to standard form, the
skills in completing the square that you have learned in your lower
mathematics will come very useful.
Completing the square in an expression like 𝑥 2 + 4𝑥 means
determining the term to be added that will produce a perfect
polynomial square. We can see that the coefficient of 𝑥 2 is 1, so we
only have to take the half of coefficient of 𝑥 which is 4, divide it by
two that gives 2 and take the square of it which 4. Then we obtain
the expression𝑥 2 + 4𝑥 + 4 = (𝑥 + 2)2 , a perfect square.
How about this example?
3𝑥 2 + 12𝑥
3(𝑥 2 + 4𝑥) factoring out 3
3(𝑥 2 + 4𝑥 + 4) taking half of 4 and squaring it.
3(𝑥 + 2)2 obtaining a perfect square

10
When completing square in an equation, we have to bear in mind
that any extra term introduced on one side must be also added to
the other side.

Let us now do some exercises. Find the center and the radius of the
following equation of a circle. Sketch the graph and indicate the
center.
(1) 𝑥 2 + 𝑦 2 − 6𝑥 = 7
(2) 𝑥 2 + 𝑦 2 − 14𝑥 + 2𝑦 = −14
(3) 16𝑥 2 + 16𝑦 2 + 96𝑥 + 40𝑦 = 315
Solution for number 1
𝑥 2 + 𝑦 2 − 6𝑥 = 7 Rewrite the equation on
𝑥 2 − 6𝑥 + 9 + 𝑦 2 = 7 + 9 standard form
(𝑥 2 − 6𝑥 + 9) + 𝑦 2 = 16
(𝑥 − 3)2 + 𝑦 2 = 16 Determine the center and radius
Center (3,0), r= 4
Graph :

11
Solution for number 2
𝑥 2 + 𝑦 2 − 14𝑥 + 2𝑦 = −14 Rewrite the equation on
𝑥 2 − 14𝑥 + 𝑦 2 + 2𝑦 = −14 standard form
(𝑥 2 − 14𝑥 + 49) + (𝑦 2 + 2𝑦 + 1) = −14 + 49 + 1
(𝑥 − 7)2 + (𝑦 + 1)2 = 36 Determine the center and radius
Center (7,-1), r= 6
Graph :

Solution for number 3

16𝑥 2 + 16𝑦 2 + 96𝑥 − 40𝑦 = 315 Rewrite the equation on
16𝑥 2 + 96𝑥 + 16𝑦 2 − 40𝑦 = 315 standard form
5
16(𝑥 2 + 6𝑥) + 16(𝑦 2 − 𝑦) = 315
2
5 25 25
16(𝑥 2 + 6𝑥 + 9) + 16 (𝑦 2 − 𝑦 + ) = 315 + 16(9) + (16)
2 16 16

2
5 2
16(𝑥 + 3) + 16 (𝑦 − ) = 484
4

2
5 2 484 121 11 2
(𝑥 + 3) + (𝑦 − ) = = =( )
4 16 4 2
12
 2
5 2 11 2 Determine the center and
(𝑥 + 3) + (𝑦 − ) = ( )
4 2 radius
5 11
Center (-3, − ), 𝑟 =
4 2

Graph :

Problems Involving Circles

Let us now explore at some real life situation where we can use our
knowledge involving circle.

Problem 1. A street with two lanes, each 10 ft wide goes through a

semi-circular tunnel with radius 12 ft. How high is the a the edge of
each lane? Round off to 2 decimal places.

13
 Figure 6
Solution. We draw a coordinate system with origin at the middle of
the highway, as shown in Figure 6. Because of the given radius, the
tunnel’s boundary is on the circle x2 + y2 = 122. Point P is the point on
the arc just above the edge of a lane, so its x-coordinate is 10. We
need its y-coordinate. We then solve 102 + y2 = 122 for y > 0, giving us
y = 2√11 or 6.63 ft.

Problem 2. Tikboy dropped a stone into a pond creating a circular

ripple. The radius of the ripple is increased by 4cm/s. Help him define
an equation that models the circular ripple 10 seconds after the
stone is dropped.

14
Solution.
𝑥2 + 𝑦2 = 𝑟2 The stone entered the water (0,0) ,
4𝑐𝑚 determine the radius of a circle,
𝑟=( ) (10 𝑠)
𝑠
multiply the rate with the time
= 40 𝑐𝑚
𝑥 2 + 𝑦 2 = 402 Substituting the value of r
𝑥 2 + 𝑦 2 = 1600 Equation of the circular ripple

V. Activities

A. Check your understanding

Directions: Read carefully and apply what you have learned
about the properties of circle.
1. The graph below shows a circle with the center (0,0)
(a) What is the x intercept
(b) What is the y intercept
(c) What is the radius
(d) Write the equation of the circle

Fig 7 : Graph for Question 1

15
 2. In each item, give the standard equation of the circle satisfying
the given conditions.
(a) center at the origin contains (0,-3)
(b) center at (-1, -5) with a diameter of 8
(c) center (-2,-3) tangent to the y axis
(d) center (-2, -3) tangent to the x axis
(e) contains the point (-2,0) and (8,0) the radius is 5
3. A circle has its center at (0.0) and passes through the point (8,5)
(a) Calculate the radius of the circle.
(b) Write the equation of the circle.
(c) Sketch the graph.

B. Converting General to Standard Form

Directions: Convert the following equation of a circle into its
standard form. Locate the center and find the radius.
1. 𝑥 2 + 𝑦 2 + 8𝑦 = 33
2. 4𝑥 2 + 4𝑦 2 − 16𝑥 + 40𝑦 − 67 = 0
3. 4𝑥 2 + 12𝑥 + 4𝑦 2 − 16𝑦 − 11 = 0
4. 𝑥 2 − 4𝑥 + 𝑦 2 − 4𝑦 − 8 = 0
5. 𝑥 2 + 𝑦 2 − 14𝑥 + 12𝑦 = 36
6. 𝑥 2 + 10𝑥 + 𝑦 2 − 16𝑦 − 11 = 0
7. 4𝑥 2 + 4𝑦 2 = 25
8. 2𝑥 2 + 2𝑦 2 − 14𝑥 + 18𝑦 = 7
9. 16𝑥 2 + 80𝑥 + 16𝑦 2 − 112𝑦 + 247 = 0
10. 𝑥 2 + 𝑦 2 − 10𝑦 + 25 = 0

16
C. Word Problem.

Directions: Solve the following word problem.

(1) Two satellites are orbiting Earth. The path of one satellite has
the equation 𝑥 2 + 𝑦 2 = 56 250 000. The orbit of the satellite is
200 km farther from the center of Earth. In one orbit, how
much farther does the second satellite travel than the first
satellite?

(2) A circular play area with radius 3m is to be partitioned using

a straight fence as shown in figure 8 how long should the
fence be?

Figure 8. partitioning using straight fence

(3) A ferris wheel is elevated 1 m above ground. When a car

reaches the highest point on the ferris wheel, its altitude from
ground level is 31 m. How far away from the center,
horizontally, is the car when it is at an altitude of 25 m?

17
 VI. Assessment

A. Check your understanding

Directions: Read carefully and apply what you have learned
about the properties of circle.
1. The graph below shows a circle with the center (3,-3)
(e) What is the x intercept
(f) What is the y intercept
(g) What is the radius
(h) Write the equation of the circle

Fig 9 : Graph for question 1

4. In each item, write an equation of a circle that models each

situation. Assume that the origin (0,0) is the center of the circle.

18
 (a) The possible location of a distress airplane 11miles away
from the control tower
(b) The rim of motorcycle with a diameter of 121 cm
(c) The base of ‘bibingka’ clay pan with an area of 64 𝜋 in2
(d) The path of a satellite in a circular orbit at a distance of 10
000 km from the center of the Earth.
(e) The ring that has a diameter of 2𝜋 𝑐𝑚

B. Converting General to Standard Form

Directions: Convert the following equation of a circle into its

standard form. Locate the center and find the radius.

1. 4𝑥 2 + 4𝑦 2 − 20𝑥 + 40𝑦 + 5 = 0
2. 9𝑥 2 + 9𝑦 2 + 42𝑥 + 84𝑦 + 65 = 0
3. 𝑥 2 + 𝑦 2 − 2𝑥 − 6𝑦 = −6
4. 𝑥 2 − 2𝑥 + 𝑦 2 + 4𝑦 − 31 = 0
5. 𝑥 2 + 𝑦 2 + 10𝑥 = −25
6. 𝑥 2 − 12𝑥 + 84 = −𝑦 2 + 16𝑦
7. 𝑥 2 + 𝑦 2 − 4𝑥 + 4𝑦 = 1
8. 4𝑥 2 + 4𝑦 2 + 32𝑦 − 36 = 0
9. 16𝑥 2 + 80𝑥 + 16𝑦 2 − 112𝑦 = −24
10. 𝑥 2 + 𝑦 2 + 14𝑥 + 10𝑦 + 62 = 0

19
C. Word Problem.
Directions: Solve the following word problem.

(1) A window is to be constructed as shown, with its upper

boundary the arc of a circle having radius 4 ft and center at
the midpoint of base AD. If the vertical side is to be 3/4 as
long as the base, find the dimensions (vertical side and
base) of this window. Round off your final answer to two
decimal places.

(2) A waterway in the theme park has a semicircular cross

section with diameter 11 feet. The boats that are going to be
used in the waterway have rectangular cross sections and
are found to submerge 1 feet into the water. If the waterway
is to be filled with water 4.5 deep what is the maximum
possible width of the boat. Refer to Figure 9.

Figure 9

20
(3) A stone is dropped into a pond, creating a circular ripple.
The Radius of the ripple increase progressively at 5cm/s. A
toy boat is floating on the pond, 3 m east and 4 m north of
the point where the stone is dropped. How long does the
ripple take to reach the toy boat?

VII. Reflection

Directions: Fill in the blanks with appropriate words, phrase or

expression to make the sentences complete and sensible.

(1) Circle is a conic section in which the plane cuts the

cone ____________________.
(2) The __________________ form of a circle is a point.
(3) Circle is a special type of ______________. In which the
coefficient of x2 is equal to the coefficient of _______.

(4) There are four types of conic sections; they are circle,
__________, ellipse, and ________________.

(5) To solve for the length of the radius we use the

__________ formula.

(6) The degenerate form of parabola is ________________.

(7) The general form of the equation (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 =

𝑟 2 is __________________.

(8) The center of the circle (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 is ______.

(9) The radius of the circle (𝑥 − 2)2 + (𝑦 − 3)2 = 5 is ______.

(10) The graph of (𝑥 − 2)2 + (𝑦 − 1)2 = 1 is

21
VIII. Challenge Yourself (Early Finishers/Differentiation)

A truck with a wide load, proceeding slowly along a SCTX

Expressway is approaching the Subic Tunnel that is shaped
like a semi-circle. The maximum height of the tunnel is 5.25 m.
If the load is 8m wide and 3.5m high, will it fit through the
tunnel? Explain your reasoning.

22
IX. References

Chris Kirkpatrick. Principle of Mathematics 10, 2010. Toronto,

Canada: Nelson Education Ltd. 88-93

Joy Ascano, Martin J. Jr, Olofernes, A., Tolentino M.A.

PreCalculus Learners Material, 2016. Pasig City,
Philippines: DepEd-BLR. 7-12

James Stewart, Redlin L., Watson S. PreCalculus Mathematics

For Calculus 7 ed. 2016. Boston MA: Cengage
Learning. 790-793

Lumencandela. n.d. “Introduction to Conics Section” Boundless

Algebra. Accessed July 19, 2020. https://courses.
lumenlearning.com/boundless-
algebra/chapter/introduction-to-conic-sections/

Lumencandela. n.d. “Rotation of Axes” Algebra and

Trigonometry. Accessed July 19, 2020
https://courses.lumenlearning.com/suny-
osalgebratrig/chapter/rotation-of-axes/

Prepared by:

Rommel S. Daz
Special Science Teacher I
Zambales NHS - Iba, District

23
X. Key to Corrections

Activities:
A. Check Your Understanding
1. (a) (3,0), (-3,0)
(b) (0,3), (0,-3)
(c) 3
(d) 𝑥 2 + 𝑦 2 = 9
2. (a) 𝑥 2 + 𝑦 2 = 9
(b) (𝑥 + 1)2 + (𝑦 + 5)2 = 16
(c) (𝑥 + 2)2 + (𝑦 + 3)2 = 9
(d) (𝑥 + 2)2 + (𝑦 + 3)2 = 4
(e) (𝑥 − 3)2 + 𝑦 2 = 25
3. (a) 5
(b) 𝑥 2 + 𝑦 2 = 25
(c)

B. General to Standard Form

1. 𝑥 2 + (𝑦 + 4)2 = 49 C (0,-4) r=7
49 7
2. (𝑥 − 2)2 + (𝑦 + 5)2 = C (2,-5) r=
4 2
3 2 3
3. (𝑥 + ) + 2
(𝑦 − 2) = 3 C (− , 2) r=3
2 2
4. (𝑥 − 2) + (𝑦 − 2) = 2
2 2
C (2, 2) r=4
5. (𝑥 − 7)2 + (𝑦 + 6)2 = 121 C (7,-6) r=11
6. (𝑥 + 5)2 + (𝑦 − 8)2 = 49 C (-5,8) r=10

24
 25 5
7. 𝑥 2 + 𝑦 2 = C (0, 0) r=
4 2
5 2 9 2
8. (𝑥 − ) + (𝑦 + ) = 36 C (0,-4) r=6
2 2
9. (𝑥 − 6) + (𝑦 − 5) = 49
2 2
C (6, 5) r=7
10. degenerate circle C (0, 5) r=0

C. General to Standard Form

(1) 1257 Km
(2) 4√2 m
(3) 12 m
The ferris wheel, must be considered to be 1 unit above the x axis
(ground level), center on the y axis, and highest point at y = 31.
The diameter is thus 30, and the radius 15. We locate the center at
(0, 16), and write the equation of the circle as x2 + (y − 16)2 = 152.
If y = 25, we have x2 + (25−16)2 = 152, so x2 = 152 − 92 = 144, and x =
±12. (Clearly, there are two points on the ferris wheel at an altitude
of 25 m.) Thus, the car is 12 m away horizontally from the center.

Assessment:
A. Check Your Understanding
2. (a) (3,0)
(b) (0,-3)
(c) 3
(d) (𝑥 − 3)2 + (𝑦 = 3)2 = 9
2. (a) 𝑥 2 + 𝑦 2 = 121
(b) 𝑥 2 + 𝑦 2 = 81
(c) 𝑥 2 + 𝑦 2 = 64
(d) 𝑥 2 + 𝑦 2 = 100 000 000
(e) 𝑥 2 + 𝑦 2 = 1

B. General to Standard Form

5 5
1. (𝑥 − )2 + (𝑦 + 5)2 = 30 C ( ,-5) r=√30
2 2
7 2 14 2
2. (𝑥 + C (− 3,− 3 ) r=2√5
7 14
) + (𝑦 + ) = 40
3 3
3. (𝑥 − 1) + (𝑦 + 3) = 4
2 2
C (1,-3) r=2

25
 4. (𝑥 − 1)2 + (𝑦 + 2)2 = 36 C (1,-2) r=6
5. Degenerate circle C (-5, 0) r=0
6. (𝑥 − 6)2 + (𝑦 − 8)2 = 16 C (6, 8) r=4
7. (𝑥 − 2)2 + (𝑦 + 2)2 = 9 C (2, -2) r=3
8. 𝑥 2 + (𝑦 + 4)2 = 25 C (0, -4) r=5
5 7 5 7
9. (𝑥 − )2 + (𝑦 + )2 = 20 C ( ,− ) r=2√5
2 2 2 2
10. (𝑥 + 7)2 + (𝑦 + 5)2 = 10 C (6, 8) r=√10

C. Word Problem
(1) base 4.44 ft, side 3.33 ft
(2) 10.42 ft
(3) 37 sec

Reflection:
Check Your Understanding
(1) horizontally
(2) degenerate
(3) ellipse, y2
(4) parabola, hyperbola
(5) distance
(6) two intersecting lines
(7) 𝑥 2 + 𝑦 2 + 𝐵𝑥 + 𝐶𝑥 + 𝐷 = 0
(8) (h, k)
(9) √5
(10)

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